How do use the first derivative test to determine the local extrema #3-x^2#?

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Answer 1 · 2015-08-30T15:07:26

Refer Explanation section

Given -

#y = 3 - x^2#

#dy/dx = - 2x#

#dy/dx = 0 => - 2x = 0#

#x = 0#

At #x = 0#, #y = 3 - (0^2) = 3#

#(d^2y)/dx^2 = - 2 < 0#

As the second derivative is less than zero , the function has a maximum at (0, 3)

graph{3 - x^2 [-10, 10, -5, 5]}